What is the integral from 1 to 6 of 4/x-4/(x^2) ?

The integral from 1 to 6 of 4/x-4/(x^2) is 4ln(6)-10/3

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integral from 1 to 6 of 4/x-4/(x^2)

\int_{1}^{6} \frac{4}{x} - \frac{4}{x ^{2}} d x

4 ln (6) - \frac{10}{3}

Decimal

3.83370 \dots

Solution steps

\int_{1}^{6} \frac{4}{x} - \frac{4}{x ^{2}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{6} \frac{4}{x} d x - \int_{1}^{6} \frac{4}{x ^{2}} d x

\int_{1}^{6} \frac{4}{x} d x = 4 ln (6)

\int_{1}^{6} \frac{4}{x ^{2}} d x = \frac{10}{3}

= 4 ln (6) - \frac{10}{3}

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What is the integral from 1 to 6 of 4/x-4/(x^2) ?
The integral from 1 to 6 of 4/x-4/(x^2) is 4ln(6)-10/3